this is “solutions”, chapter 9 from the book introduction to chemistry

This is “Solutions”, chapter 9 from the book Introduction to Chemistry: General, Organic, and Biological(index.html) (v. 1.0).

This book is licensed under a Creative Commons by-nc-sa 3.0 (http://creativecommons.org/licenses/by-nc-sa/3.0/) license. See the license for more details, but that basically means you can share this book as long as youcredit the author (but see below), don't make money from it, and do make it available to everyone else under thesame terms.

This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz(http://lardbucket.org) in an effort to preserve the availability of this book.

Normally, the author and publisher would be credited here. However, the publisher has asked for the customaryCreative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally,per the publisher's request, their name has been removed in some passages. More information is available on thisproject's attribution page (http://2012books.lardbucket.org/attribution.html?utm_source=header).

For more information on the source of this book, or why it is available for free, please see the project's home page(http://2012books.lardbucket.org/). You can browse or download additional books there.

i

www.princexml.com

Prince - Non-commercial License

This document was created with Prince, a great way of getting web content onto paper.

index.html

index.html

http://creativecommons.org/

http://creativecommons.org/licenses/by-nc-sa/3.0/

http://creativecommons.org/licenses/by-nc-sa/3.0/

http://lardbucket.org

http://lardbucket.org

http://2012books.lardbucket.org/attribution.html?utm_source=header

http://2012books.lardbucket.org/

http://2012books.lardbucket.org/

Chapter 9

Solutions

Opening Essay

If you watch any of the medical dramas on television, you may have heard a doctor (actually an actor) call for anintravenous solution of “Ringer’s lactate” (or “lactated Ringer’s”). So what is Ringer’s lactate?

Intravenous (IV) solutions are administered for two main reasons: (1) to introduce necessary substances into thebloodstream, such as ions for proper body function, sugar and other food substances for energy, or drugs totreat a medical condition, and (2) to increase the volume of the bloodstream. Many people with acute or long-term medical conditions have received some type of an IV solution.

One basic IV solution, called a normal saline solution, is simply a dilute solution of NaCl dissolved in water. Normalsaline is 9.0 g of NaCl dissolved in each liter of solution. (The reason for this particular concentration isexplained in Section 9.4 "Properties of Solutions".)

Ringer’s lactate is a normal saline solution that also has small amounts of potassium and calcium ions mixed in. Inaddition, it contains about 2.5 g of lactate ions (C3H5O3

−) per liter of solution. The liver metabolizes lactate ions

into bicarbonate (HCO3−) ions, which help maintain the acid-base balance of blood. (Acids and bases are

discussed in Chapter 10 "Acids and Bases".) Many medical problems, such as heart attacks and shock, affect theacid-base balance of blood, and the presence of lactate in the IV solution eases problems caused by thisimbalance.

Physicians can select from a range of premade IV solutions, in accordance with a patient’s particular needs.Ringer’s lactate is commonly used when a patient’s blood volume must be increased quickly. Another frequentlyused IV solution, called D5W, is a 5% solution of dextrose (a form of sugar) in water.

Solutions are all around us. Air, for example, is a solution. If you live near a lake, ariver, or an ocean, that body of water is not pure H2O but most probably a solution.

Much of what we drink—for example, soda, coffee, tea, and milk—is at least in part asolution. Solutions are a large part of everyday life.

431

A lot of the chemistry occurring around us happens in solution. In fact, much of thechemistry that occurs in our own bodies takes place in solution, and manysolutions—such as the Ringer’s lactate IV solution—are important for our health. Inour understanding of chemistry, we need to understand a little bit about solutions.In this chapter, you will learn about the special characteristics of solutions, howsolutions are characterized, and some of their properties.

Chapter 9 Solutions

432

9.1 Solutions

LEARNING OBJECTIVE

1. Understand what causes solutions to form.

A solution1 is another name for a homogeneous mixture. Chapter 1 "Chemistry,Matter, and Measurement" defined a mixture as a material composed of two or moresubstances. In a solution, the combination is so intimate that the differentsubstances cannot be differentiated by sight, even with a microscope. Compare, forexample, a mixture of salt and pepper and another mixture consisting of salt andwater. In the first mixture, we can readily see individual grains of salt and the flecksof pepper. A mixture of salt and pepper is not a solution. However, in the secondmixture, no matter how carefully we look, we cannot see two different substances.Salt dissolved in water is a solution.

The major component of a solution, called the solvent2, is typically the same phaseas the solution itself. Each minor component of a solution (and there may be morethan one) is called the solute3. In most of the solutions we will describe in thistextbook, there will be no ambiguity about whether a component is the solvent orthe solute.) For example, in a solution of salt in water, the solute is salt, and solventis water.

Solutions come in all phases, and the solvent and the solute do not have to be in thesame phase to form a solution (such as salt and water). For example, air is a gaseoussolution of about 80% nitrogen and about 20% oxygen, with some other gasespresent in much smaller amounts. An alloy4 is a solid solution consisting of a metal(like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy ofiron and carbon and small amounts of other metals, is an example of a solidsolution. Table 9.1 "Types of Solutions" lists some common types of solutions, withexamples of each.

Table 9.1 Types of Solutions

Solvent Phase Solute Phase Example

gas gas air

liquid gas carbonated beverages

1. Another name for ahomogeneous mixture.

2. The major component of asolution.

3. The minor component of asolution.

4. A solid solution of a metal withother substances dissolved init.

Chapter 9 Solutions

433

Solvent Phase Solute Phase Example

liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages)

liquid solid saltwater

solid gas H2 gas absorbed by Pd metal

solid liquid Hg(ℓ) in dental fillings

solid solid steel alloys

What causes a solution to form? The simple answer is that the solvent and thesolute must have similar intermolecular interactions. When this is the case, theindividual particles of solvent and solute can easily mix so intimately that eachparticle of solute is surrounded by particles of solute, forming a solution. However,if two substances have very different intermolecular interactions, large amounts ofenergy are required to force their individual particles to mix intimately, so asolution does not form.

This process leads to a simple rule of thumb: like dissolves like. Solvents that are verypolar will dissolve solutes that are very polar or even ionic. Solvents that arenonpolar will dissolve nonpolar solutes. Thus water, being polar, is a good solventfor ionic compounds and polar solutes like ethanol (C2H5OH). However, water does

not dissolve nonpolar solutes, such as many oils and greases (Figure 9.1"Solubility").

Chapter 9 Solutions

9.1 Solutions 434

Figure 9.1 Solubility

Because of different intermolecular interactions, oil (on top) and water (bottom, colored red) do not dissolve in eachother.

© Thinkstock

We use the word soluble5 to describe a solute that dissolves in a particular solvent,and the word insoluble6 for a solute that does not dissolve in a solvent. Thus, wesay that sodium chloride is soluble in water but insoluble in hexane (C6H14). If the

solute and the solvent are both liquids and soluble in any proportion, we use theword miscible7, and the word immiscible8 if they are not.

5. A solute that dissolves in aparticular solvent.

6. A solute that does not dissolvein a particular solvent.

7. Liquids that dissolve in eachother.

8. Liquids that do not dissolve ineach other.

Chapter 9 Solutions

9.1 Solutions 435

EXAMPLE 1

Water is considered a polar solvent. Which substances should dissolve inwater?

1. methanol (CH3OH)2. sodium sulfate (Na2SO4)3. octane (C8H18)

Solution

Because water is polar, substances that are polar or ionic will dissolve in it.

1. Because of the OH group in methanol, we expect its molecules to bepolar. Thus, we expect it to be soluble in water. As both water andmethanol are liquids, the word miscible can be used in place of soluble.

2. Sodium sulfate is an ionic compound, so we expect it to be soluble inwater.

3. Like other hydrocarbons, octane is nonpolar, so we expect that it wouldnot be soluble in water.

SKILL-BUILDING EXERCISE

Toluene (C6H5CH3) is widely used in industry as a nonpolar solvent. Whichsubstances should dissolve in toluene?

1. water (H2O)

2. sodium sulfate (Na2SO4)

3. octane (C8H18)

CONCEPT REVIEW EXERCISES

1. What causes a solution to form?

2. How does the phrase like dissolves like relate to solutions?

Chapter 9 Solutions

9.1 Solutions 436

ANSWERS

1. Solutions form because a solute and a solvent have similar intermolecularinteractions.

2. It means that substances with similar intermolecular interactions will dissolvein each other.

KEY TAKEAWAY

• Solutions form because a solute and a solvent experience similarintermolecular interactions.

Chapter 9 Solutions

9.1 Solutions 437

EXERCISES

1. Define solution.

2. Give several examples of solutions.

3. What is the difference between a solvent and a solute?

4. Can a solution have more than one solute in it? Can you give an example?

5. Does a solution have to be a liquid? Give several examples to support youranswer.

6. Give at least two examples of solutions found in the human body.

7. Which substances will probably be soluble in water, a very polar solvent?

a. sodium nitrate (NaNO3)b. hexane (C6H14)c. isopropyl alcohol [(CH3)2CHOH]d. benzene (C6H6)

8. Which substances will probably be soluble in toluene (C6H5CH3), a nonpolarsolvent?

a. sodium nitrate (NaNO3)b. hexane (C6H14)c. isopropyl alcohol [(CH3)2CHOH]d. benzene (C6H6)

9. The solubility of alcohols in water varies with the length of carbon chain. Forexample, ethanol (CH3CH2OH) is soluble in water in any ratio, while only0.0008 mL of heptanol (CH3CH2CH2CH2CH2CH2CH2OH) will dissolve in 100 mLof water. Propose an explanation for this behavior.

10. Dimethyl sulfoxide [(CH3)2SO] is a polar liquid. Based on the information inExercise 9, which do you think will be more soluble in it—ethanol or heptanol?

Chapter 9 Solutions

9.1 Solutions 438

ANSWERS

1. a homogeneous mixture

3. A solvent is the majority component of a solution; a solute is the minoritycomponent of a solution.

5. A solution does not have to be liquid; air is a gaseous solution, while somealloys are solid solutions (answers will vary).

7. a. probably solubleb. probably not solublec. probably solubled. probably not soluble

9. Small alcohol molecules have strong polar intermolecular interactions, so theydissolve in water. In large alcohol molecules, the nonpolar end overwhelms thepolar end, so they do not dissolve very well in water.

Chapter 9 Solutions

9.1 Solutions 439

9.2 Concentration

LEARNING OBJECTIVES

1. Express the amount of solute in a solution in various concentrationunits.

2. Use molarity to determine quantities in chemical reactions.3. Determine the resulting concentration of a diluted solution.

To define a solution precisely, we need to state its concentration9: how muchsolute is dissolved in a certain amount of solvent. Words such as dilute orconcentrated are used to describe solutions that have a little or a lot of dissolvedsolute, respectively, but these are relative terms whose meanings depend onvarious factors.

Solubility

There is usually a limit to how much solute will dissolve in a given amount ofsolvent. This limit is called the solubility10 of the solute. Some solutes have a verysmall solubility, while other solutes are soluble in all proportions. Table 9.2"Solubilities of Various Solutes in Water at 25°C (Except as Noted)" lists thesolubilities of various solutes in water. Solubilities vary with temperature, so Table9.2 "Solubilities of Various Solutes in Water at 25°C (Except as Noted)" includes thetemperature at which the solubility was determined.

Table 9.2 Solubilities of Various Solutes in Water at 25°C (Except as Noted)

Substance Solubility (g in 100 mL of H2O)

AgCl(s) 0.019

C6H6(ℓ) (benzene) 0.178

CH4(g) 0.0023

CO2(g) 0.150

CaCO3(s) 0.058

CaF2(s) 0.0016

Ca(NO3)2(s) 143.9

9. How much solute is dissolvedin a certain amount of solvent.

10. The limit of how much solutecan be dissolved in a givenamount of solvent.

Chapter 9 Solutions

440

Substance Solubility (g in 100 mL of H2O)

C6H12O6 (glucose) 120.3 (at 30°C)

KBr(s) 67.8

MgCO3(s) 2.20

NaCl(s) 36.0

NaHCO3(s) 8.41

C12H22O11 (sucrose) 204.0 (at 20°C)

If a solution contains so much solute that its solubility limit is reached, the solutionis said to be saturated11, and its concentration is known from informationcontained in Table 9.2 "Solubilities of Various Solutes in Water at 25°C (Except asNoted)". If a solution contains less solute than the solubility limit, it isunsaturated12. Under special circumstances, more solute can be dissolved evenafter the normal solubility limit is reached; such solutions are called supersaturatedand are not stable. If the solute is solid, excess solute can easily recrystallize. If thesolute is a gas, it can bubble out of solution uncontrollably, like what happens whenyou shake a soda can and then immediately open it.

Note

Recrystallization of excess solute from a supersaturated solution usually givesoff energy as heat. Commercial heat packs containing supersaturated sodiumacetate (NaC2H3O2) take advantage of this phenomenon. You can probably find

them at your local drugstore.

Most solutions we encounter are unsaturated, so knowing the solubility of thesolute does not accurately express the amount of solute in these solutions. Thereare several common ways of specifying the concentration of a solution.

Percent Composition

There are several ways of expressing the concentration of a solution by using apercentage. The mass/mass percent13 (% m/m) is defined as the mass of a solutedivided by the mass of a solution times 100:

11. A solution whose solute is at itssolubility limit.

12. A solution whose solute is lessthan its solubility limit.

13. A concentration unit thatrelates the mass of the soluteto the mass of the solution.

Chapter 9 Solutions

9.2 Concentration 441

If you can measure the masses of the solute and the solution, determining themass/mass percent is easy. Each mass must be expressed in the same units todetermine the proper concentration.

EXAMPLE 2

A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. Whatis the mass/mass percent concentration of the solution?

Solution

We can substitute the quantities given in the equation for mass/masspercent:


1. A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 ghas 15.8 g of dextrose dissolved in it. What is the mass/mass percentconcentration of the solution?

For gases and liquids, volumes are relatively easy to measure, so the concentrationof a liquid or a gas solution can be expressed as a volume/volume percent14 (% v/v): the volume of a solute divided by the volume of a solution times 100:

Again, the units of the solute and the solution must be the same. A hybridconcentration unit, mass/volume percent15 (% m/v), is commonly used forintravenous (IV) fluids (Figure 9.2 "Mass/Volume Percent"). It is defined as themass in grams of a solute, divided by volume in milliliters of solution times 100:

% m/m =mass of solute

mass of solution× 100%

% m/m =36.5 g355 g

× 100% = 10.3%

% v/v =volume of solute

volume of solution× 100%

14. A concentration unit thatrelates the volume of the soluteto the volume of the solution.

15. A concentration unit thatrelates the mass of the soluteto the volume of the solution.

Chapter 9 Solutions


Figure 9.2 Mass/VolumePercent

The 0.9% concentration on this IVbag is mass/volume percent.

© Thinkstock

Each percent concentration can be used to produce aconversion factor between the amount of solute, theamount of solution, and the percent. Furthermore,given any two quantities in any percent composition,the third quantity can be calculated, as the followingexample illustrates.

EXAMPLE 3

A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of115 mL. What volume of ethanol solute does the sample contain?

Solution

A percentage concentration is simply the number of parts of solute per 100parts of solution. Thus, the percent concentration of 45.0% v/v implies thefollowing:

That is, there are 45 mL of C2H5OH for every 100 mL of solution. We can usethis fraction as a conversion factor to determine the amount of C2H5OH in115 mL of solution:

% m/v =mass of solute (g)

volume of solution (mL)× 100%

45.0% v/v →45 mL C2H5OH100 mL solution

115 mL solution ×45 mL C2H5OH

100 mL solution= 51.8 mL C2H5OH

Chapter 9 Solutions


Note

The highest concentration of ethanol that can be obtained normally is 95%ethanol, which is actually 95% v/v.


1. What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is neededto obtain 50.0 g of C6H12O6?

EXAMPLE 4

A normal saline IV solution contains 9.0 g of NaCl in every liter of solution.What is the mass/volume percent of normal saline?

Solution

We can use the definition of mass/volume percent, but first we have toexpress the volume in milliliter units:

1 L = 1,000 mL

Because this is an exact relationship, it does not affect the significant figuresof our result.


1. The chlorine bleach that you might find in your laundry room is typicallycomposed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mLof solution. What is the mass/volume percent of the bleach?

% m/v =9.0 g NaCl

1,000 mL solution× 100% = 0.90% m/v

Chapter 9 Solutions


In addition to percentage units, the units for expressing the concentration ofextremely dilute solutions are parts per million (ppm)16 and parts per billion(ppb)17. Both of these units are mass based and are defined as follows:

Note

Similar to parts per million and parts per billion, related units include parts perthousand (ppth) and parts per trillion (ppt).

Concentrations of trace elements in the body—elements that are present inextremely low concentrations but are nonetheless necessary for life—are commonlyexpressed in parts per million or parts per billion. Concentrations of poisons andpollutants are also described in these units. For example, cobalt is present in thebody at a concentration of 21 ppb, while the State of Oregon’s Department ofAgriculture limits the concentration of arsenic in fertilizers to 9 ppm.

Note

In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb isequivalent to 1 µg/L.

ppm =mass of solute

mass of solution× 1,000,000

ppb =mass of solute

mass of solution× 1,000,000,000

16. The mass of a solute comparedto the mass of a solution times1,000,000.

17. The mass of a solute comparedto the mass of a solution times1,000,000,000.

Chapter 9 Solutions


EXAMPLE 5

If the concentration of cobalt in a human body is 21 ppb, what mass in gramsof Co is present in a body having a mass of 70.0 kg?

Solution

A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g ofsolution.” Written as a conversion factor, this concentration of Co is asfollows:

We can use this as a conversion factor, but first we must convert 70.0 kg togram units:

Now we determine the amount of Co:

This is only 1.5 mg.


1. An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in partsper million?

Molarity

Another way of expressing concentration is to give the number of moles of soluteper unit volume of solution. Such concentration units are useful for discussing

21 ppb Co →21 g Co

1,000,000,000 g solution

70.0 kg ×1,000 g

1 kg= 7.00 × 104 g

7.00 × 104 g solution ×21 g Co

1,000,000,000 g solution= 0.0015 g Co

Chapter 9 Solutions


chemical reactions in which a solute is a product or a reactant. Molar mass can thenbe used as a conversion factor to convert amounts in moles to amounts in grams.

Molarity18 is defined as the number of moles of a solute dissolved per liter ofsolution:

Molarity is abbreviated M (often referred to as “molar”), and the units are oftenabbreviated as mol/L. It is important to remember that “mol” in this expressionrefers to moles of solute and that “L” refers to liters of solution. For example, if youhave 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore

which is read as “three point oh molar sodium chloride.” Sometimes (aq) is addedwhen the solvent is water, as in “3.0 M NaCl(aq).”

Before a molarity concentration can be calculated, the amount of the solute must beexpressed in moles, and the volume of the solution must be expressed in liters, asdemonstrated in the following example.

molarity =number of moles of solute

number of liters of solution

1.5 mol NaCl0.500 L solution

= 3.0 M NaCl

18. Number of moles of solute perliter of solution.

Chapter 9 Solutions


EXAMPLE 6

What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?

Solution

Before we substitute these quantities into the definition of molarity, wemust convert them to the proper units. The mass of NaOH must beconverted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol:

Next, we convert the volume units from milliliters to liters:

Now that the quantities are expressed in the proper units, we can substitutethem into the definition of molarity:


1. If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is themolarity of this caffeine solution?

The definition of molarity can also be used to calculate a needed volume of solution,given its concentration and the number of moles desired, or the number of moles ofsolute (and subsequently, the mass of the solute), given its concentration andvolume. The following example illustrates this.

25.0 g NaOH ×1 mol NaOH

40.00 g NaOH= 0.625 mol NaOH

750 mL ×1 L

1,000 mL= 0.750 L

M =0.625 mol NaOH

0.750 L= 0.833 M NaOH

Chapter 9 Solutions


EXAMPLE 7

1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] isneeded to obtain 0.450 mol of the compound?

2. Ethylene glycol (C2H6O2) is mixed with water to make auto enginecoolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueoussolution?

Solution

In both parts, we will use the definition of molarity to solve for the desiredquantity.

1.

To solve for the volume of solution, we multiply both sides byvolume of solution and divide both sides by the molarity value toisolate the volume of solution on one side of the equation:

Note that because the definition of molarity is mol/L, thedivision of mol by M yields L, a unit of volume.

2. The molar mass of C2H6O2 is 62.08 g/mol., so

To solve for the number of moles of solute, we multiply bothsides by the volume:

moles of solute = (6.00 M)(5.00 L) = 30.0 mol

Note that because the definition of molarity is mol/L, theproduct M × L gives mol, a unit of amount. Now, using the molarmass of C3H8O3, we convert mol to g:

0.0753 M =0.450 mol (CH 3)2NH

volume of solution

volume of solution =0.450 mol (CH 3)2NH

0.0753 M= 5.98 L

6.00 M =moles of solute

5.00 L

Chapter 9 Solutions


Thus, there are 1,860 g of C2H6O2 in the specified amount ofengine coolant.

Note

Dimethylamine has a “fishy” odor. In fact, organic compounds called aminescause the odor of decaying fish. (For more information about amines, seeChapter 15 "Organic Acids and Bases and Some of Their Derivatives", Section15.1 "Functional Groups of the Carboxylic Acids and Their Derivatives" andSection 15.11 "Amines: Structures and Names" through Section 15.13 "Aminesas Bases".)


1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain0.888 mol of HCOOH?

2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 arein 0.565 L of a 0.955 M solution?

Using Molarity in Stoichiometry Problems

Of all the ways of expressing concentration, molarity is the one most commonlyused in stoichiometry problems because it is directly related to the mole unit.Consider the following chemical equation:

HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

Suppose we want to know how many liters of aqueous HCl solution will react with agiven mass of NaOH. A typical approach to answering this question is as follows:

30.0 mol ×62.08 g

mol= 1,860 g

Chapter 9 Solutions


In itself, each step is a straightforward conversion. It is the combination of the stepsthat is a powerful quantitative tool for problem solving.

Chapter 9 Solutions


EXAMPLE 8

How many milliliters of a 2.75 M HCl solution are needed to react with 185 gof NaOH? The balanced chemical equation for this reaction is as follows:

HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

Solution

We will follow the flowchart to answer this question. First, we convert themass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol:

Using the balanced chemical equation, we see that there is a one-to-oneratio of moles of HCl to moles of NaOH. We use this to determine the numberof moles of HCl needed to react with the given amount of NaOH:

Finally, we use the definition of molarity to determine the volume of 2.75 MHCl needed:

We need 1,680 mL of 2.75 M HCl to react with the NaOH.

185 g NaOH ×1 mol NaOH

40.00 g NaOH= 4.63 mol NaOH

4.63 mol NaOH ×1 mol HCl

1 mol NaOH= 4.63 mol HCl

2.75 M HCl =4.63 mol HCl

volume of HCl solution

volume of HCl =4.63 mol HCl2.75 M HCl

= 1.68 L ×1,000 mL

1 L= 1,680 mL

Chapter 9 Solutions



1. How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 gof Ca(OH)2? The balanced chemical equation for the reaction is as follows:

H2SO4(aq) + Ca(OH)2(s) → 2H2O(ℓ) + CaSO4(aq)

The general steps for performing stoichiometry problems such as this are shown inFigure 9.3 "Diagram of Steps for Using Molarity in Stoichiometry Calculations". Youmay want to consult this figure when working with solutions in chemical reactions.The double arrows in Figure 9.3 "Diagram of Steps for Using Molarity inStoichiometry Calculations" indicate that you can start at either end of the chartand, after a series of simple conversions, determine the quantity at the other end.

Figure 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations

When using molarity in stoichiometry calculations, a specific sequence of steps usually leads you to the correctanswer.

Chapter 9 Solutions


Many of the fluids found in our bodies are solutions. The solutes range from simpleionic compounds to complex proteins. Table 9.3 "Approximate Concentrations ofVarious Solutes in Some Solutions in the Body*" lists the typical concentrations ofsome of these solutes.

Table 9.3 Approximate Concentrations of Various Solutes in Some Solutions in theBody*

Solution Solute Concentration (M)

Na+ 0.138

K+ 0.005

Ca2+ 0.004

Mg2+ 0.003

Cl− 0.110

blood plasma

HCO3− 0.030

stomach acid HCl 0.10

NaCl 0.15

PO43− 0.05urine

NH2CONH2 (urea) 0.30

*Note: Concentrations are approximate and can vary widely.

Chapter 9 Solutions


Looking Closer: The Dose Makes the Poison

Why is it that we can drink 1 qt of water when we are thirsty and not beharmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying:the dose makes the poison. This means that what may be dangerous in someamounts may not be dangerous in other amounts.

Take arsenic, for example. Some studies show that arsenic deprivation limitsthe growth of animals such as chickens, goats, and pigs, suggesting that arsenicis actually an essential trace element in the diet. Humans are constantlyexposed to tiny amounts of arsenic from the environment, so studies ofcompletely arsenic-free humans are not available; if arsenic is an essential tracemineral in human diets, it is probably required on the order of 50 ppb or less. Atoxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over140 times the trace amount that may be required by the body. Thus, arsenic isnot poisonous in and of itself. Rather, it is the amount that is dangerous: thedose makes the poison.

Similarly, as much as water is needed to keep us alive, too much of it is alsorisky to our health. Drinking too much water too fast can lead to a conditioncalled water intoxication, which may be fatal. The danger in water intoxicationis not that water itself becomes toxic. It is that the ingestion of too much watertoo fast dilutes sodium ions, potassium ions, and other salts in the bloodstreamto concentrations that are not high enough to support brain, muscle, and heartfunctions. Military personnel, endurance athletes, and even desert hikers aresusceptible to water intoxication if they drink water but do not replenish thesalts lost in sweat. As this example shows, even the right substances in thewrong amounts can be dangerous!

Chapter 9 Solutions


Is this athlete poisoning himself? Drinking too much water too fast can actually be dangerous—even fatal.

© Thinkstock

Equivalents

Concentrations of ionic solutes are occasionally expressed in units calledequivalents (Eq)19. One equivalent equals 1 mol of positive or negative charge.Thus, 1 mol/L of Na+(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/Lsolution of Ca2+(aq) ions has a concentration of 2 Eq/L because calcium has a 2+charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example,human blood plasma has a total concentration of about 150 mEq/L. (For moreinformation about the ions present in blood plasma, see Chapter 3 "Ionic Bondingand Simple Ionic Compounds", Section 3.3 "Formulas for Ionic Compounds".)

Dilution

When solvent is added to dilute a solution, the volume of the solution changes, butthe amount of solute does not change. Before dilution, the amount of solute wasequal to its original concentration times its original volume:

amount in moles = (concentration × volume)initial

After dilution, the same amount of solute is equal to the final concentration timesthe final volume:19. One mole of charge (either

positive or negative).

Chapter 9 Solutions


amount in moles = (concentration × volume)final

To determine a concentration or amount after a dilution, we can use the followingequation:

(concentration × volume)initial = (concentration × volume)final

Any units of concentration and volume can be used, as long as both concentrationsand both volumes have the same unit.

EXAMPLE 9

A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the finalconcentration of the diluted solution?

Solution

Because the volume units are the same, and we are looking for the molarityof the final solution, we can use (concentration × volume)initial =(concentration × volume)final:

(0.900 M × 125 mL) = (concentration × 1,125 mL)

We solve by isolating the unknown concentration by itself on one side of theequation. Dividing by 1,125 mL gives

as the final concentration.


1. A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin isan anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. Whatis the concentration of the resulting heparin solution?

concentration =0.900 M × 125 mL

1,125 mL= 0.100 M

Chapter 9 Solutions



1. What are some of the units used to express concentration?

2. Distinguish between the terms solubility and concentration.

ANSWERS

1. % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)

2. Solubility is typically a limit to how much solute can dissolve in a givenamount of solvent. Concentration is the quantitative amount of solutedissolved at any concentration in a solvent.

KEY TAKEAWAYS

• Various concentration units are used to express the amounts of solute ina solution.

• Concentration units can be used as conversion factors in stoichiometryproblems.

• New concentrations can be easily calculated if a solution is diluted.

Chapter 9 Solutions


EXERCISES

1. Define solubility. Do all solutes have the same solubility?

2. Explain why the terms dilute or concentrated are of limited usefulness indescribing the concentration of solutions.

3. If the solubility of sodium chloride (NaCl) is 30.6 g/100 mL of H2O at a giventemperature, how many grams of NaCl can be dissolved in 250.0 mL of H2O?

4. If the solubility of glucose (C6H12O6) is 120.3 g/100 mL of H2O at a giventemperature, how many grams of C6H12O6 can be dissolved in 75.0 mL of H2O?

5. How many grams of sodium bicarbonate (NaHCO3) can a 25.0°C saturatedsolution have if 150.0 mL of H2O is used as the solvent?

6. If 75.0 g of potassium bromide (KBr) are dissolved in 125 mL of H2O, is thesolution saturated, unsaturated, or supersaturated?

7. Calculate the mass/mass percent of a saturated solution of NaCl. Use the datafrom Table 9.2 "Solubilities of Various Solutes in Water at 25°C (Except asNoted)", assume that masses of the solute and the solvent are additive, and usethe density of H2O (1.00 g/mL) as a conversion factor.

8. Calculate the mass/mass percent of a saturated solution of MgCO3 Use the datafrom Table 9.2 "Solubilities of Various Solutes in Water at 25°C (Except asNoted)", assume that masses of the solute and the solvent are additive, and usethe density of H2O (1.00 g/mL) as a conversion factor.

9. Only 0.203 mL of C6H6 will dissolve in 100.000 mL of H2O. Assuming that thevolumes are additive, find the volume/volume percent of a saturated solutionof benzene in water.

10. Only 35 mL of aniline (C6H5NH2) will dissolve in 1,000 mL of H2O. Assumingthat the volumes are additive, find the volume/volume percent of a saturatedsolution of aniline in water.

11. A solution of ethyl alcohol (C2H5OH) in water has a concentration of 20.56% v/v. What volume of C2H5OH is present in 255 mL of solution?

12. What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution?

13. The average human body contains 5,830 g of blood. What mass of arsenic ispresent in the body if the amount in blood is 0.55 ppm?

Chapter 9 Solutions


14. The Occupational Safety and Health Administration has set a limit of 200 ppmas the maximum safe exposure level for carbon monoxide (CO). If an averagebreath has a mass of 1.286 g, what is the maximum mass of CO that can beinhaled at that maximum safe exposure level?

15. Which concentration is greater—15 ppm or 1,500 ppb?

16. Express the concentration 7,580 ppm in parts per billion.

17. What is the molarity of 0.500 L of a potassium chromate solution containing0.0650 mol of K2CrO4?

18. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea[(NH2)2CO]?

19. What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr?

20. If 3.08 g of Ca(OH)2 is dissolved in enough water to make 0.875 L of solution,what is the molarity of the Ca(OH)2?

21. What mass of HCl is present in 825 mL of a 1.25 M solution?

22. What mass of isopropyl alcohol (C3H8O) is dissolved in 2.050 L of a 4.45 Maqueous C3H8O solution?

23. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?

24. How many milliliters of a 0.0015 M cocaine hydrochloride (C17H22ClNO4)solution is needed to obtain 0.010 g of the solute?

25. Aqueous calcium chloride reacts with aqueous silver nitrate according to thefollowing balanced chemical equation:

CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)

How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl2 react withexcess AgNO3? How many grams of AgCl are made?

26. Sodium bicarbonate (NaHCO3) is used to react with acid spills. The reactionwith sulfuric acid (H2SO4) is as follows:

2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)

If 27.6 mL of a 6.25 M H2SO4 solution were spilled, how many moles of NaHCO3would be needed to react with the acid? How many grams of NaHCO3 is this?

27. The fermentation of glucose to make ethanol and carbon dioxide has thefollowing overall chemical equation:

C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)

Chapter 9 Solutions


If 1.00 L of a 0.567 M solution of C6H12O6 were completely fermented, whatwould be the resulting concentration of the C2H5OH solution? How manymoles of CO2 would be formed? How many grams is this? If each mole of CO2had a volume of 24.5 L, what volume of CO2 is produced?

28. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:

2NaHSO3(aq) → Na2SO3(aq) + H2O(ℓ) + SO2(g)

If 567 mL of a 1.005 M NaHSO3 solution were heated until all the NaHSO3 hadreacted, what would be the resulting concentration of the Na2SO3 solution?How many moles of SO2 would be formed? How many grams of SO2 would beformed? If each mole of SO2 had a volume of 25.78 L, what volume of SO2would be produced?

29. What is the concentration of a 1.0 M solution of K+(aq) ions in equivalents/liter?

30. What is the concentration of a 1.0 M solution of SO42−(aq) ions in equivalents/

liter?

31. A solution having initial concentration of 0.445 M and initial volume of 45.0 mLis diluted to 100.0 mL. What is its final concentration?

32. A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is itsfinal mass/volume percent?

Chapter 9 Solutions


ANSWERS

1. Solubility is the amount of a solute that can dissolve in a given amount ofsolute, typically 100 mL. The solubility of solutes varies widely.

3. 76.5 g

5. 12.6 g

7. 26.5%

9. 0.203%

11. 52.4 mL

13. 0.00321 g

15. 15 ppm

17. 0.130 M

19. 0.208 M

21. 37.6 g

23. 0.496 L

25. 0.262 mol; 37.5 g

27. 1.13 M C2H5OH; 1.13 mol of CO2; 49.7 g of CO2; 27.7 L of CO2

29. 1.0 Eq/L

31. 0.200 M

Chapter 9 Solutions


9.3 The Dissolution Process

LEARNING OBJECTIVE

1. Describe the dissolution process at the molecular level.

What occurs at the molecular level to cause a solute to dissolve in a solvent? Theanswer depends in part on the solute, but there are some similarities common to allsolutes.

Recall the rule that like dissolves like. As we saw in Section 9.1 "Solutions", thismeans that substances must have similar intermolecular forces to form solutions.When a soluble solute is introduced into a solvent, the particles of solute caninteract with the particles of solvent. In the case of a solid or liquid solute, theinteractions between the solute particles and the solvent particles are so strongthat the individual solute particles separate from each other and, surrounded bysolvent molecules, enter the solution. (Gaseous solutes already have theirconstituent particles separated, but the concept of being surrounded by solventparticles still applies.) This process is called solvation20 and is illustrated in Figure9.4 "Solvation". When the solvent is water, the word hydration21, rather thansolvation, is used.

Figure 9.4 Solvation

When a solute dissolves, the individual particles of solute become surrounded by solvent particles. Eventually theparticle detaches from the remaining solute, surrounded by solvent molecules in solution.

20. The process by which soluteparticles are surrounded bysolvent particles.

21. Solvation by water molecules.

Chapter 9 Solutions

463

Source: Photo © Thinkstock

In the case of molecular solutes like glucose, the solute particles are individualmolecules. However, if the solute is ionic, the individual ions separate from eachother and become surrounded by solvent particles. That is, the cations and anionsof an ionic solute separate when the solute dissolves. This process is referred to asdissociation22. Compare the dissociation of a simple ionic solute as shown in Figure9.5 "Ionic Dissociation" to the process illustrated in Figure 9.4 "Solvation".

Figure 9.5 Ionic Dissociation

When an ionic solute dissolves, the individual ions separate from each other as they go into solution.

Source: Photo © Thinkstock

The dissociation of soluble ionic compounds gives solutions of these compounds aninteresting property: they conduct electricity. Because of this property, solubleionic compounds are referred to as electrolytes23. Many ionic compoundsdissociate completely and are therefore called strong electrolytes24. Sodiumchloride is an example of a strong electrolyte. Some compounds dissolve butdissociate only partially, and solutions of such solutes may conduct electricity onlyweakly. These solutes are called weak electrolytes25. Acetic acid (CH3COOH), the

compound in vinegar, is a weak electrolyte. Solutes that dissolve into individualneutral molecules without dissociation do not impart additional electricalconductivity to their solutions and are called nonelectrolytes26. Table sugar(C12H22O11) is an example of a nonelectrolyte.

22. The process of cations andanions of an ionic soluteseparating when the solutedissolves.

23. An ionic compound thatdissolves in water.

24. An ionic compound thationizes completely when itdissolves.

25. An ionic compound that doesnot ionize completely when itdissolves.

26. A compound that does notionize at all when it dissolves.

Chapter 9 Solutions

9.3 The Dissolution Process 464

Note

The term electrolyte is used in medicine to mean any of the important ions thatare dissolved in aqueous solution in the body. Important physiologicalelectrolytes include Na+, K+, Ca2+, Mg2+, and Cl−.

EXAMPLE 10

The following substances all dissolve to some extent in water. Classify eachas an electrolyte or a nonelectrolyte.

1. potassium chloride (KCl)2. fructose (C6H12O6)3. isopropyl alcohol [CH3CH(OH)CH3]4. magnesium hydroxide [Mg(OH)2]

Solution

Each substance can be classified as an ionic solute or a nonionic solute. Ionicsolutes are electrolytes, and nonionic solutes are nonelectrolytes.

1. Potassium chloride is an ionic compound; therefore, when it dissolves,its ions separate, making it an electrolyte.

2. Fructose is a sugar similar to glucose. (In fact, it has the same molecularformula as glucose.) Because it is a molecular compound, we expect it tobe a nonelectrolyte.

3. Isopropyl alcohol is an organic molecule containing the alcoholfunctional group. The bonding in the compound is all covalent, so whenisopropyl alcohol dissolves, it separates into individual molecules butnot ions. Thus, it is a nonelectrolyte.

4. Magnesium hydroxide is an ionic compound, so when it dissolves itdissociates. Thus, magnesium hydroxide is an electrolyte.

Chapter 9 Solutions


Note

More information than that provided in this chapter is needed to determine ifsome electrolytes are strong or weak. We will consider this in Chapter 10 "Acidsand Bases".


The following substances all dissolve to some extent in water. Classify eachas an electrolyte or a nonelectrolyte.

1. acetone (CH3COCH3)

2. iron(III) nitrate [Fe(NO3)3]

3. elemental bromine (Br2)

4. sodium hydroxide (NaOH)

CONCEPT REVIEW EXERCISE

1. Explain how the solvation process describes the dissolution of a solute in asolvent.

ANSWER

1. Each particle of the solute is surrounded by particles of the solvent, carryingthe solute from its original phase.

KEY TAKEAWAY

• When a solute dissolves, its individual particles are surrounded bysolvent molecules and are separated from each other.

Chapter 9 Solutions


EXERCISES

1. Describe what happens when an ionic solute like Na2SO4 dissolves in a polarsolvent.

2. Describe what happens when a molecular solute like sucrose (C12H22O11)dissolves in a polar solvent.

3. Classify each substance as an electrolyte or a nonelectrolyte. Each substancedissolves in H2O to some extent.

a. NH4NO3b. CO2c. NH2CONH2d. HCl

4. Classify each substance as an electrolyte or a nonelectrolyte. Each substancedissolves in H2O to some extent.

a. CH3CH2CH2OHb. Ca(CH3CO2)2c. I2d. KOH

5. Will solutions of each solute conduct electricity when dissolved?

a. AgNO3b. CHCl3c. BaCl2d. Li2O

6. Will solutions of each solute conduct electricity when dissolved?

a. CH3COCH3b. N(CH3)3c. CH3CO2C2H5d. FeCl2

Chapter 9 Solutions


ANSWERS

1. Each ion of the ionic solute is surrounded by particles of solvent, carrying theion from its associated crystal.

3. a. electrolyteb. nonelectrolytec. nonelectrolyted. electrolyte

5. a. yesb. noc. yesd. yes

Chapter 9 Solutions


9.4 Properties of Solutions

LEARNING OBJECTIVE

1. Describe how the properties of solutions differ from those of puresolvents.

Solutions are likely to have properties similar to those of their majorcomponent—usually the solvent. However, some solution properties differsignificantly from those of the solvent. Here, we will focus on liquid solutions thathave a solid solute, but many of the effects we will discuss in this section areapplicable to all solutions.

Colligative Properties

Solutes affect some properties of solutions that depend only on the concentration ofthe dissolved particles. These properties are called colligative properties27. Fourimportant colligative properties that we will examine here are vapor pressuredepression, boiling point elevation, freezing point depression, and osmoticpressure.

Molecular compounds separate into individual molecules when they are dissolved,so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast,ionic compounds separate into their constituent ions when they dissolve, so 1 molof an ionic compound will produce more than 1 mol of dissolved particles. Forexample, every mole of NaCl that dissolves yields 1 mol of Na+ ions and 1 mol of Cl−

ions, for a total of 2 mol of particles in solution. Thus, the effect on a solution’sproperties by dissolving NaCl may be twice as large as the effect of dissolving thesame amount of moles of glucose (C6H12O6).

Vapor Pressure Depression

All liquids evaporate. In fact, given enough volume, a liquid will turn completelyinto a vapor. If enough volume is not present, a liquid will evaporate only to thepoint where the rate of evaporation equals the rate of vapor condensing back into aliquid. The pressure of the vapor at this point is called the vapor pressure28 of theliquid.

27. A characteristic of solutionsthat depends only on thenumber of dissolved particles.

28. The pressure of a vapor that isin equilibrium with its liquidphase.

Chapter 9 Solutions

469

The presence of a dissolved solid lowers the characteristic vapor pressure of a liquidso that it evaporates more slowly. (The exceptions to this statement are if the soluteitself is a liquid or a gas, in which case the solute will also contribute something tothe evaporation process. We will not discuss such solutions here.) This property iscalled vapor pressure depression29 and is depicted in Figure 9.6 "Vapor PressureDepression".

Figure 9.6 Vapor Pressure Depression

The presence of solute particles blocks some of the ability for liquid particles to evaporate. Thus, solutions of solidsolutes typically have a lower vapor pressure than the pure solvent.

Boiling Point and Freezing Point Effects

A related property of solutions is that their boiling points are higher than theboiling point of the pure solvent. Because the presence of solute particles decreasesthe vapor pressure of the liquid solvent, a higher temperature is needed to reachthe boiling point. This phenomenon is called boiling point elevation30. For everymole of particles dissolved in a liter of water, the boiling point of water increases byabout 0.5°C.29. The lowering of the vapor

pressure of a solution versusthe pure solvent.

30. The raising of the boiling pointof a solution versus the puresolvent.

Chapter 9 Solutions

9.4 Properties of Solutions 470

Note

Some people argue that putting a pinch or two of salt in water used to cookspaghetti or other pasta makes a solution that has a higher boiling point, so thepasta cooks faster. In actuality, the amount of solute is so small that the boilingpoint of the water is practically unchanged.

The presence of solute particles has the opposite effect on the freezing point of asolution. When a solution freezes, only the solvent particles come together to forma solid phase, and the presence of solute particles interferes with that process.Therefore, for the liquid solvent to freeze, more energy must be removed from thesolution, which lowers the temperature. Thus, solutions have lower freezing pointsthan pure solvents do. This phenomenon is called freezing point depression31. Forevery mole of particles in a liter of water, the freezing point decreases by about1.9°C.

Both boiling point elevation and freezing point depression have practical uses. Forexample, solutions of water and ethylene glycol (C2H6O2) are used as coolants in

automobile engines because the boiling point of such a solution is greater than100°C, the normal boiling point of water. In winter, salts like NaCl and CaCl2 are

sprinkled on the ground to melt ice or keep ice from forming on roads andsidewalks (Figure 9.7 "Effect of Freezing Point Depression"). This is because thesolution made by dissolving sodium chloride or calcium chloride in water has alower freezing point than pure water, so the formation of ice is inhibited.

31. The lowering of the freezingpoint of a solution versus thepure solvent.

Chapter 9 Solutions


Figure 9.7 Effect of Freezing Point Depression

The salt sprinkled on this sidewalk makes the water on the sidewalk have a lower freezing point than pure water, soit does not freeze as easily. This makes walking on the sidewalk less hazardous in winter.

© Thinkstock

EXAMPLE 11

Which solution’s freezing point deviates more from that of pure water—a 1M solution of NaCl or a 1 M solution of CaCl2?

Solution

Colligative properties depend on the number of dissolved particles, so thesolution with the greater number of particles in solution will show thegreatest deviation. When NaCl dissolves, it separates into two ions, Na+ andCl−. But when CaCl2 dissolves, it separates into three ions—one Ca2+ ion andtwo Cl− ions. Thus, mole for mole, CaCl2 will have 50% more impact onfreezing point depression than NaCl.

Chapter 9 Solutions



1. Which solution’s boiling point deviates more from that of pure water—a 1 Msolution of CaCl2 or a 1 M solution of MgSO4?

Osmotic Pressure

The last colligative property of solutions we will consider is a very important onefor biological systems. It involves osmosis32, the process by which solventmolecules can pass through certain membranes but solute particles cannot. Whentwo solutions of different concentration are present on either side of thesemembranes (called semipermeable membranes), there is a tendency for solventmolecules to move from the more dilute solution to the more concentrated solutionuntil the concentrations of the two solutions are equal. This tendency is calledosmotic pressure33. External pressure can be exerted on a solution to counter theflow of solvent; the pressure required to halt the osmosis of a solvent is equal to theosmotic pressure of the solution.

Osmolarity34 (osmol) is a way of reporting the total number of particles in asolution to determine osmotic pressure. It is defined as the molarity of a solutetimes the number of particles a formula unit of the solute makes when it dissolves(represented by i):

osmol = M × i

If more than one solute is present in a solution, the individual osmolarities areadditive to get the total osmolarity of the solution. Solutions that have the sameosmolarity have the same osmotic pressure. If solutions of differing osmolarities arepresent on opposite sides of a semipermeable membrane, solvent will transfer fromthe lower-osmolarity solution to the higher-osmolarity solution. Counterpressureexerted on the high-osmolarity solution will reduce or halt the solvent transfer. Aneven higher pressure can be exerted to force solvent from the high-osmolaritysolution to the low-osmolarity solution, a process called reverse osmosis. Reverseosmosis is used to make potable water from saltwater where sources of fresh waterare scarce.

32. The process by which solventmolecules can pass throughcertain membranes but soluteparticles cannot.

33. The tendency for solventmolecules to move from themore dilute solution to themore concentrated solutionuntil the concentrations of thetwo solutions are equal.

34. A way of reporting the totalnumber of particles in asolution to determine theosmotic pressure.

Chapter 9 Solutions


EXAMPLE 12

A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO3)2 aqueous solution areplaced on opposite sides of a semipermeable membrane. Determine theosmolarity of each solution and predict the direction of solvent flow.

Solution

The solvent will flow into the solution of higher osmolarity. The NaCl soluteseparates into two ions—Na+ and Cl−—when it dissolves, so its osmolarity isas follows:

osmol (NaCl) = 0.50 M × 2 = 1.0 osmol

The Ca(NO3)2 solute separates into three ions—one Ca2+ and two NO3−—when

it dissolves, so its osmolarity is as follows:

osmol [Ca(NO3)2] = 0.30 M × 3 = 0.90 osmol

The osmolarity of the Ca(NO3)2 solution is lower than that of the NaClsolution, so water will transfer through the membrane from the Ca(NO3)2

solution to the NaCl solution.


1. A 1.5 M C6H12O6 aqueous solution and a 0.40 M Al(NO3)3 aqueous solution areplaced on opposite sides of a semipermeable membrane. Determine theosmolarity of each solution and predict the direction of solvent flow.

Chapter 9 Solutions


To Your Health: Dialysis

The main function of the kidneys is to filter the blood to remove wastes andextra water, which are then expelled from the body as urine. Some diseases robthe kidneys of their ability to perform this function, causing a buildup of wastematerials in the bloodstream. If a kidney transplant is not available ordesirable, a procedure called dialysis can be used to remove waste materialsand excess water from the blood.

In one form of dialysis, called hemodialysis, a patient’s blood is passed though alength of tubing that travels through an artificial kidney machine (also called adialysis machine). A section of tubing composed of a semipermeable membraneis immersed in a solution of sterile water, glucose, amino acids, and certainelectrolytes. The osmotic pressure of the blood forces waste molecules andexcess water through the membrane into the sterile solution. Red and whiteblood cells are too large to pass through the membrane, so they remain in theblood. After being cleansed in this way, the blood is returned to the body.

A patient undergoinghemodialysis depends on osmosisto cleanse the blood of wasteproducts that the kidneys areincapable of removing due todisease.

© Thinkstock

Chapter 9 Solutions


Dialysis is a continuous process, as the osmosis of waste materials and excesswater takes time. Typically, 5–10 lb of waste-containing fluid is removed ineach dialysis session, which can last 2–8 hours and must be performed severaltimes a week. Although some patients have been on dialysis for 30 or moreyears, dialysis is always a temporary solution because waste materials areconstantly building up in the bloodstream. A more permanent solution is akidney transplant.

Cell walls are semipermeable membranes, so the osmotic pressures of the body’sfluids have important biological consequences. If solutions of different osmolarityexist on either side of the cells, solvent (water) may pass into or out of the cells,sometimes with disastrous results. Consider what happens if red blood cells areplaced in a hypotonic solution, meaning a solution of lower osmolarity than theliquid inside the cells. The cells swell up as water enters them, disrupting cellularactivity and eventually causing the cells to burst. This process is called hemolysis. Ifred blood cells are placed in a hypertonic solution, meaning one having a higherosmolarity than exists inside the cells, water leaves the cells to dilute the externalsolution, and the red blood cells shrivel and die. This process is called crenation.Only if red blood cells are placed in isotonic solutions that have the same osmolarityas exists inside the cells are they unaffected by negative effects of osmotic pressure.Glucose solutions of about 0.31 M, or sodium chloride solutions of about 0.16 M, areisotonic with blood plasma.

Note

The concentration of an isotonic sodium chloride (NaCl) solution is only halfthat of an isotonic glucose (C6H12O6) solution because NaCl produces two ions

when a formula unit dissolves, while molecular C6H12O6 produces only one

particle when a formula unit dissolves. The osmolarities are therefore the sameeven though the concentrations of the two solutions are different.

Osmotic pressure explains why you should not drink seawater if you are abandonedin a life raft in the middle of the ocean. Its osmolarity is about three times higherthan most bodily fluids. You would actually become thirstier as water from yourcells was drawn out to dilute the salty ocean water you ingested. Our bodies do a

Chapter 9 Solutions


better job coping with hypotonic solutions than with hypertonic ones. The excesswater is collected by our kidneys and excreted.

Osmotic pressure effects are used in the food industry to make pickles fromcucumbers and other vegetables and in brining meat to make corned beef. It is alsoa factor in the mechanism of getting water from the roots to the tops of trees!

Career Focus: Perfusionist

A perfusionist is a medical technician trained to assist during any medicalprocedure in which a patient’s circulatory or breathing functions requiresupport. The use of perfusionists has grown rapidly since the advent of open-heart surgery in 1953.

Most perfusionists work in operating rooms, where their main responsibility isto operate heart-lung machines. During many heart surgeries, the heart itselfmust be stopped. In these situations, a heart-lung machine keeps the patientalive by aerating the blood with oxygen and removing carbon dioxide. Theperfusionist monitors both the machine and the status of the blood, notifyingthe surgeon and the anesthetist of any concerns and taking corrective action ifthe status of the blood becomes abnormal.

Despite the narrow parameters of their specialty, perfusionists must be highlytrained. Certified perfusion education programs require a student to learnanatomy, physiology, pathology, chemistry, pharmacology, math, and physics.A college degree is usually required. Some perfusionists work with otherexternal artificial organs, such as hemodialysis machines and artificial livers.


1. What are the colligative properties of solutions?

2. Explain how the following properties of solutions differ from those of the puresolvent: vapor pressure, boiling point, freezing point, and osmotic pressure.

Chapter 9 Solutions


ANSWERS

1. Colligative properties are characteristics that a solution has that depend onthe number, not the identity, of solute particles.

2. In solutions, the vapor pressure is lower, the boiling point is higher, thefreezing point is lower, and the osmotic pressure is higher.

KEY TAKEAWAY

• Certain properties of solutions differ from those of pure solvents inpredictable ways.

Chapter 9 Solutions


EXERCISES

1. In each pair of aqueous systems, which will have the lower vapor pressure?

a. pure water or 1.0 M NaClb. 1.0 M NaCl or 1.0 M C6H12O6c. 1.0 M CaCl2 or 1.0 M (NH4)3PO4

2. In each pair of aqueous systems, which will have the lower vapor pressure?

a. 0.50 M Ca(NO3)2 or 1.0 M KBrb. 1.5 M C12H22O11 or 0.75 M Ca(OH)2c. 0.10 M Cu(NO3)2 or pure water

3. In each pair of aqueous systems, which will have the higher boiling point?

a. pure water or a 1.0 M NaClb. 1.0 M NaCl or 1.0 M C6H12O6c. 1.0 M CaCl2 or 1.0 M (NH4)3PO4

4. In each pair of aqueous systems, which will have the higher boiling point?

a. 1.0 M KBrb. 1.5 M C12H22O11 or 0.75 M Ca(OH)2c. 0.10 M Cu(NO3)2 or pure water

5. Estimate the boiling point of each aqueous solution. The boiling point of purewater is 100.0°C.

a. 0.50 M NaClb. 1.5 M Na2SO4c. 2.0 M C6H12O6

6. Estimate the freezing point of each aqueous solution. The freezing point ofpure water is 0.0°C.

a. 0.50 M NaClb. 1.5 M Na2SO4c. 2.0 M C6H12O6

7. Explain why salt (NaCl) is spread on roads and sidewalks to inhibit iceformation in cold weather.

8. Salt (NaCl) and calcium chloride (CaCl2) are used widely in some areas tominimize the formation of ice on sidewalks and roads. One of these ionic

Chapter 9 Solutions


compounds is better, mole for mole, at inhibiting ice formation. Which is thatlikely to be? Why?

9. What is the osmolarity of each aqueous solution?

a. 0.500 M NH2CONH2b. 0.500 M NaBrc. 0.500 M Ca(NO3)2

10. What is the osmolarity of each aqueous solution?

a. 0.150 M KClb. 0.450 M (CH3)2CHOHc. 0.500 M Ca3(PO4)2

11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol.What can you conclude about the salt?

12. A 1.5 M NaCl solution and a 0.75 M Al(NO3)3 solution exist on opposite sides ofa semipermeable membrane. Determine the osmolarity of each solution andthe direction of solvent flow, if any, across the membrane.

Chapter 9 Solutions


ANSWERS

1. a. 1.0 M NaClb. 1.0 M NaClc. 1.0 M (NH4)3PO4

3. a. 1.0 M NaClb. 1.0 M NaClc. 1.0 M (NH4)3PO4

5. a. 100.5°Cb. 102.3°Cc. 101°C

7. NaCl lowers the freezing point of water, so it needs to be colder for the waterto freeze.

9. a. 0.500 osmolb. 1.000 osmolc. 1.500 osmol

11. It must separate into three ions when it dissolves.

Chapter 9 Solutions


9.5 End-of-Chapter Material

Chapter 9 Solutions

482

Chapter Summary

To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in thefollowing summary and ask yourself how they relate to the topics in the chapter.

A solution is a homogeneous mixture. The major component is the solvent, while the minor component is thesolute. Solutions can have any phase; for example, an alloy is a solid solution. Solutes are soluble or insoluble,meaning they dissolve or do not dissolve in a particular solvent. The terms miscible and immiscible, instead ofsoluble and insoluble, are used for liquid solutes and solvents. The statement like dissolves like is a useful guide topredicting whether a solute will dissolve in a given solvent.

The amount of solute in a solution is represented by the concentration of the solution. The maximum amountof solute that will dissolve in a given amount of solvent is called the solubility of the solute. Such solutions aresaturated. Solutions that have less than the maximum amount are unsaturated. Most solutions areunsaturated, and there are various ways of stating their concentrations. Mass/mass percent, volume/volumepercent, and mass/volume percent indicate the percentage of the overall solution that is solute. Parts permillion (ppm) and parts per billion (ppb) are used to describe very small concentrations of a solute. Molarity,defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistrylaboratory. Equivalents express concentrations in terms of moles of charge on ions. When a solution is diluted,we use the fact that the amount of solute remains constant to be able to determine the volume or concentrationof the final diluted solution.

Dissolving occurs by solvation, the process in which particles of a solvent surround the individual particles of asolute, separating them to make a solution. For water solutions, the word hydration is used. If the solute ismolecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from eachother, forming a solution that conducts electricity. Such solutions are called electrolytes. If the dissociation ofions is complete, the solution is a strong electrolyte. If the dissociation is only partial, the solution is a weakelectrolyte. Solutions of molecules do not conduct electricity and are called nonelectrolytes.

Solutions have properties that differ from those of the pure solvent. Some of these are colligative properties,which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligativeproperties include vapor pressure depression, boiling point elevation, freezing point depression, andosmotic pressure. Osmotic pressure is particularly important in biological systems. It is caused by osmosis, thepassage of solvents through certain membranes like cell walls. The osmolarity of a solution is the product of asolution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can bereversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in someparts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose functionthrough either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffectedbecause osmotic pressure is equal on either side of the cell membrane.

Chapter 9 Solutions

9.5 End-of-Chapter Material 483

ADDITIONAL EXERCISES

1. Calcium nitrate reacts with sodium carbonate to precipitate solid calciumcarbonate:

Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + NaNO3(aq)

a. Balance the chemical equation.b. How many grams of Na2CO3 are needed to react with 50.0 mL of 0.450 M

Ca(NO3)2?c. Assuming that the Na2CO3 has a negligible effect on the volume of the

solution, find the osmolarity of the NaNO3 solution remaining after theCaCO3 precipitates from solution.

2. The compound HCl reacts with sodium carbonate to generate carbon dioxidegas:

HCl(aq) + Na2CO3(aq) → H2O(ℓ) + CO2(g) + NaCl(aq)

a. Balance the chemical equation.b. How many grams of Na2CO3 are needed to react with 250.0 mL of 0.755 M

HCl?c. Assuming that the Na2CO3 has a negligible effect on the volume of the

solution, find the osmolarity of the NaCl solution remaining after thereaction is complete.

3. Estimate the freezing point of concentrated aqueous HCl, which is usually soldas a 12 M solution. Assume complete ionization into H+ and Cl− ions.

4. Estimate the boiling point of concentrated aqueous H2SO4, which is usuallysold as an 18 M solution. Assume complete ionization into H+ and HSO4

− ions.

5. Seawater can be approximated by a 3.0% m/m solution of NaCl in water.Determine the molarity and osmolarity of seawater. Assume a density of 1.0 g/mL.

6. Human blood can be approximated by a 0.90% m/m solution of NaCl in water.Determine the molarity and osmolarity of blood. Assume a density of 1.0 g/mL.

7. How much water must be added to 25.0 mL of a 1.00 M NaCl solution to make aresulting solution that has a concentration of 0.250 M?

8. Sports drinks like Gatorade are advertised as capable of resupplying the bodywith electrolytes lost by vigorous exercise. Find a label from a sports drinkcontainer and identify the electrolytes it contains. You should be able toidentify several simple ionic compounds in the ingredients list.

Chapter 9 Solutions


9. Occasionally we hear a sensational news story about people stranded in alifeboat on the ocean who had to drink their own urine to survive. Whiledistasteful, this act was probably necessary for survival. Why not simply drinkthe ocean water? (Hint: See Exercise 5 and Exercise 6 above. What wouldhappen if the two solutions in these exercises were on opposite sides of asemipermeable membrane, as we would find in our cell walls?)

ANSWERS

1. a. Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)b. 2.39 gc. 1.80 osmol

3. −45.6°C

5. 0.513 M; 1.026 osmol

7. 75.0 mL

9. The osmotic pressure of seawater is too high. Drinking seawater would causewater to go from inside our cells into the more concentrated seawater,ultimately killing the cells.

Chapter 9 Solutions


this is “solutions”, chapter 9 from the book introduction to chemistry

Documents